Question

If $m$ and $n$ are the roots of the equation $(x+p)(x+q)-k=0$, then the roots of the equation $(x-m)(x-n)+k=0$ are-

• A. $p$ and $q$
• B. $1/p$ and $1/q$
• C. $-p$ and $-q$
• D. $p+q$ and $p-q$

Answer 1

Answer: (C)

$-p$ and $-q$

$(x+p)(x+q)-k=0⟹x^2+(p+q)x+pq-k=0$

$m$ and $n$ are the roots of this equation

So, we have

Sum of roots $=-(p+q)=m+n$

Product of the roots $=pq-k=mn$

$\Rightarrow pq=mn+k$

Consider, $(x-m)(x-n)+k=0$

$\Rightarrow x^2-(m+n)x+mn+k=0$

Sum of roots is $m+n$

But $m+n=(-p)+(-q)$

Product of the roots $=mn+k$

But $mn+k=pq=(-p)(-q)$

Hence, the roots of the new equation are $-p,-q$